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Question Number 59861 by ANTARES VY last updated on 15/May/19

$$\sqrt{\mathit{a}+\mathit{b}\sqrt{\mathit{c}}}=\sqrt{\frac{\mathit{a}+\sqrt{{\mathit{a}}^2-{\mathit{b}}^2\mathit{c}}}{2}}+\sqrt{\frac{\mathit{a}-\sqrt{{\mathit{a}}^2-{\mathit{b}}^2\mathit{c}}}{2}}.$$$$\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}$$

Commented by Kunal12588 last updated on 15/May/19

Commented by Kunal12588 last updated on 15/May/19

$$justput‘‘b^{2}c{}^{″}inplaceof‘‘b{}^{″}$$

Answered by tanmay last updated on 15/May/19

$$\frac{a+\sqrt{a^2-b^{2}c}}{2}$$$$=\frac{2a+2\sqrt{(a+b\sqrt{c})(a-b\sqrt{c})}}{4}$$$$=\frac{a+b\sqrt{c}+a-b\sqrt{c}+2\sqrt{(a+b\sqrt{c)(a-b\sqrt{c)}}}}{4}$$$$=\frac{{\left(\sqrt{a+b\sqrt{c}}\right)}^2+{\left(\sqrt{a-b\sqrt{c}}\right)}^2+2\sqrt{(a+b\sqrt{c})(a-b\sqrt{c)}}}{4}$$$$=\frac{{[\sqrt{a+b\sqrt{c}}+\sqrt{a-b\sqrt{c}}]}^2}{2^2}$$$$so$$$$\sqrt{\frac{a+\sqrt{a^2-b^{2}c}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b^{2}c}}{2}}$$$$=\frac{\sqrt{a+b\sqrt{c}}+\sqrt{a-b\sqrt{c}}}{2}+\frac{\sqrt{a+b\sqrt{c}}-\sqrt{a-b\sqrt{c}}}{2}$$$$=\sqrt{a+b\sqrt{c}}$$$$$$

Answered by Kunal12588 last updated on 15/May/19

$$let\sqrt{a+b\sqrt{c}}=\sqrt{m}+\sqrt{n}\left(1\right)$$$$\Rightarrow a+\sqrt{b^{2}c}=m+n+2\sqrt{mn}$$$$\Rightarrow m+n=a,4mn=b^{2}c$$$$\Rightarrow m-n=\sqrt{{(m+n)}^2-4mn}=\sqrt{a^2-b^{2}c}$$$$\Rightarrow m=\frac{a+\sqrt{a^2-b^{2}c}}{2},n=\frac{a-\sqrt{a^2-b^{2}c}}{2}$$$$from\left(1\right)$$$$\sqrt{a+b\sqrt{c}}=\sqrt{m}+\sqrt{n}$$$$\Rightarrow \sqrt{\mathit{a}+\mathit{b}\sqrt{\mathit{c}}}=\sqrt{\frac{\mathit{a}+\sqrt{{\mathit{a}}^2-{\mathit{b}}^2\mathit{c}}}{2}}+\sqrt{\frac{\mathit{a}-\sqrt{{\mathit{a}}^2-{\mathit{b}}^2\mathit{c}}}{2}}$$$$\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{d}$$

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